= Solution
Put $d=\sqrt{a^2-b^2}>0$. The two roots are
$$
z_{\rm in}=-i\frac{a-d}{b},\qquad z_{\rm out}=-i\frac{a+d}{b}.
$$
Since $(a-d)(a+d)=b^2$, their moduli are $b/(a+d)<1$ and $(a+d)/b>1$. Thus exactly one root is inside the unit circle, and neither lies on it.
For the real integral use $z=e^{i\theta}$, so $\sin\theta=(z-z^{-1})/(2i)$ and $d\theta=dz/(iz)$. The <change of variables> gives
$$
\int_0^{2\pi}\frac{d\theta}{a+b\sin\theta}=\oint_{|z|=1}\frac{2\,dz}{bz^2+2iaz-b}.
$$
The contour is counterclockwise. At its single interior <simple pole>, the <residue> is
$$
\frac{2}{2bz_{\rm in}+2ia}=\frac1{id}.
$$
By the <residue theorem>, the <reciprocal affine sine integral> is therefore
$$
\boxed{\int_0^{2\pi}\frac{d\theta}{a+b\sin\theta}=2\pi i\frac1{id}=\frac{2\pi}{\sqrt{a^2-b^2}}.}
$$
The condition $a>b>0$ also makes the real denominator strictly positive, consistent with the positive answer and the absence of contour poles on the unit circle.
Back to article page
